Explanation Given data: Refer to Figure 16.60 in the textbook. Formula used: Write an expression to calculate the value of step input. u ( t ) = { 0 t < 0 1 t > 0 Write a general expression to calculate the impedance of a resistor in s -domain. Z R = R (1) Here, R is the value of resistance. Write a general expression to calculate the impedance of an inductor in s -domain. Z L = s L (2) Here, L is the value of inductance. Write a general expression to calculate the impedance of a capacitor in s -domain. Z C = 1 s C (3) Here, C is the value of capacitance. Calculation: The given circuit is redrawn as shown in Figure 1. For a DC circuit, at steady state condition at time t = 0 − the capacitor acts like open circuit and the inductor acts like short circuit. For time t < 0 : The current source i s ( t ) is calculated as follows: i s ( t ) = 20 ( 0 ) { ∵ u ( t ) = 0 for t < 0 } = 0 The voltage source v s ( t ) is calculated as follows: v s ( t ) = 20 ( 0 ) { ∵ u ( t ) = 0 for t < 0 } = 0 When the value of current source is zero, it is open circuited and when the value of voltage source is zero it is short circuited. Now, the Figure 1 is reduced as shown in Figure 2. Refer to Figure 2, there is no current and voltage source placed in the circuit. Therefore, the value of current through the inductor and capacitor is equal to zero. i L ( 0 − ) = 0 A v C ( 0 − ) = 0 V The current through inductor and voltage across capacitor is always continuous so that, i ( 0 ) = i L ( 0 − ) = i L ( 0 + ) = 0 A v ( 0 ) = v C ( 0 − ) = v C ( 0 + ) = 0 V For time t > 0 , the circuit is remains same as shown in Figure 1. Apply Laplace transform for v s ( t ) to find V s ( s ) . V s ( s ) = 20 s { ∵ L ( u ( t ) ) = 1 s } Apply Laplace transform for i s ( t ) to find I s ( s ) . I s ( s ) = 20 s { ∵ L ( u ( t ) ) = 1 s } Substitute 4 Ω for R 1 in equation (1) to find Z R 1 . Z R 1 = 4 Substitute 3 Ω for R 2 in equation (1) to find Z R 2 . Z R 2 = 3 Substitute 5 Ω for R 3 in equation (1) to find Z R 3 . Z R 3 = 5 Substitute 250 mH for L in equation (2) to find Z L . Z L = s ( 250 mH ) = s ( 250 × 10 − 3 H ) { ∵ 1 m = 10 − 3 } = 0.25 s Substitute 500 mF for C in equation (3) to find Z C . Z C = 1 s ( 500 mF ) = 1 s ( 500 × 10 − 3 F ) { ∵ 1 m = 10 − 3 } = 2 s Convert the Figure 1 into s -domain. Apply Kirchhoff’s voltage law for the circuit shown in Figure 3. 4 ( − 20 s ) + 4 I ( s ) + 3 I ( s ) + 0.25 s I ( s ) + 20 s + 5 I ( s ) + 2 s I ( s ) = 0 − 80 s + 4 I ( s ) + 3 I ( s ) + 0.25 s I ( s ) + 20 s + 5 I ( s ) + 2 s I ( s ) = 0 12 I ( s ) + 0.25 s I ( s ) − 60 s + 2 s I ( s ) = 0 [ 12 + 0.25 s + 2 s ] I ( s ) = 60 s Simplify the above equation as follows: [ 12 s + 0.25 s 2 + 2 s ] I ( s ) = 60 s [ 0.25 ( s 2 + 48 s + 8 ) s ] I ( s ) = 60 s Simplify the above equation to find I ( s ) . I ( s ) = 60 s ( s 0.25 ( s 2 + 48 s + 8 ) ) I ( s ) = 240 s 2 + 48 s + 8 (4) From the equation (4) , the characteristic equation is s 2 + 48 s + 8 = 0 (5) Write a general expression to calculate the roots of quadratic equation ( a s 2 + b s + c = 0 ) . s 1 , 2 = − b ± b 2 − 4 a c 2 a (6) Comparing the equation (5) with the equation ( a s 2 + b s + c = 0 )

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ISBN: 8220102801448

Author: Alexander

Publisher: YUZU

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Chapter 16, Problem 37P

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Find the expression of voltage v(t) for t>0 in the circuit of Figure 16.60.

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Chapter 16, Problem 36P

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Find the expression of voltage v ( t ) for t > 0 in the circuit of Figure 16.60.

Solution Summary: The author explains the expression of voltage v(t) in the circuit of Figure 16.60.

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Find the expression of voltage v(t) for t>0 in the circuit of Figure 16.60.

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Chapter 16 Solutions

Find the expression of voltage v ( t ) for t &gt; 0 in the circuit of Figure 16.60.

Solution Summary: The author explains the expression of voltage v(t) in the circuit of Figure 16.60.

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Find the expression of voltage v(t) for t>0 in the circuit of Figure 16.60.

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Chapter 16 Solutions

Find the expression of voltage v ( t ) for t > 0 in the circuit of Figure 16.60.